sum of all even numbers between 99 to 301 is?

• we have terms :100,102.....300
  tn=300,a=100,d=2
  tn=a+(n-1)d
  tn=101
  Sum=n/2(a+l)=101/2 *(100+300)
  =20200
• 10 years agoHelpfull: Yes(36) No(1)
• i think answer is 20200 bcoz there is total 101 even terms in between 99 & 301.
  so by applying formula s(n)=n/2*(a+l) i.e 101/2(100+300)=20200 where a & l are 1st and last term.
  so answer is 20200..
• 10 years agoHelpfull: Yes(11) No(1)
• Here no are: 100+102+104+.......+300
  The above sequence is in A.P where common difference(d) = 2
  first term(a) = 100
  total no of terms = 101
  so sum of sequence is n/2*(2a + (n-1)*d) = 101/2*(200+(101-1)*2) = 20200
• 10 years agoHelpfull: Yes(5) No(0)
• even terms from the given numbers will be 100,102,104.........,300
  clearly all terms are in AP whose first term is a=100 and common difference d=2, last term is l=300
  we know that l=a+(n-1)d
  300=100+(n-1)2 solving this we will get a=101
  sum of nth term is Sn=n/2(a+l)
  Sn=101/2(100+300)
  20200 this is answer
• 10 years agoHelpfull: Yes(2) No(0)
• we have AP 100,102,....,300
  a=100,d=2,tn=300
  so,n=101(apply formula-tn=a+(n-1)d) and hence
  sn=n/2*(a+tn)
  so,sum=20200
• 10 years agoHelpfull: Yes(1) No(0)
• 100,102,...................300
  let no of even term =n
  nth term=100+(n-1)2=300
  n=99
  sum=(99/2){2*100+(99-1)2=19602 ans
  
• 10 years agoHelpfull: Yes(0) No(9)
• i think 20200.
• 10 years agoHelpfull: Yes(0) No(0)
• 100+102......+300...a=100,l=300,n=51..using n/2(a+L) this formulae,ans:20200
• 10 years agoHelpfull: Yes(0) No(1)
• nth term = a+(n-1)d
  where a=first term d=difference
  hence
  300 = 100+(n-1)*2
  after solving, n=101
  sum = n/2(2*a+(n-1)*d)
  
  sum=101/2(2*100+(101-1)*2)
  
  sum = 20200.
• 10 years agoHelpfull: Yes(0) No(0)
• series will b
  100 102 104 106....300
  nw dis is the series with frst term 100 and last term 300 with common difference d= 2
  let no of terms b n
  thus,300=100+(n-1)d
  d=2
  n=101
  sn=n/2(2a+(n-1)d)
  a=100
  sn=20200
• 10 years agoHelpfull: Yes(0) No(0)